Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{-7k - 7}{-8k + 64} \times \dfrac{k^2 + k - 72}{k + 1} $
Solution: First factor the quadratic. $x = \dfrac{-7k - 7}{-8k + 64} \times \dfrac{(k - 8)(k + 9)}{k + 1} $ Then factor out any other terms. $x = \dfrac{-7(k + 1)}{-8(k - 8)} \times \dfrac{(k - 8)(k + 9)}{k + 1} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -7(k + 1) \times (k - 8)(k + 9) } { -8(k - 8) \times (k + 1) } $ $x = \dfrac{ -7(k + 1)(k - 8)(k + 9)}{ -8(k - 8)(k + 1)} $ Notice that $(k + 1)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -7\cancel{(k + 1)}(k - 8)(k + 9)}{ -8\cancel{(k - 8)}(k + 1)} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $x = \dfrac{ -7\cancel{(k + 1)}\cancel{(k - 8)}(k + 9)}{ -8\cancel{(k - 8)}\cancel{(k + 1)}} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $x = \dfrac{-7(k + 9)}{-8} $ $x = \dfrac{7(k + 9)}{8} ; \space k \neq 8 ; \space k \neq -1 $